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PHXII04:MOVING CHARGES AND MAGNETISM

362540 A Helmholtz coil has a pair of loops, each with $N$ turns and radius $R$ . They are placed coaxially at distance $R$ and the same current $I$ flows through the loops in the same direction. The magnitude of magnetic field at $P$ , midway between the centres $A$ and $C$ , is given by [Refer to figure given below]:
supporting img

1

\frac{8 N μ_{0} I}{5^{1 / 2} R}

2

\frac{4 N μ_{0} I}{5^{3 / 2} R}

3

\frac{4 N μ_{0} I}{5^{1 / 2} R}

4

\frac{8 N μ_{0} I}{5^{3 / 2} R}

Explanation:

The resultant magnetic field at $P$ is
$B = 2 (\frac{μ_{0} N I R^{2}}{2 {(R^{2} + \frac{R^{2}}{4})}^{3 / 2}}) = \frac{8 μ_{0} N I}{5^{3 / 2} R}$

JEE - 2018

PHXII04:MOVING CHARGES AND MAGNETISM

362541 $A$ and $B$ are two concentric circular loop carrying current $i_{1}$ and $i_{2}$ , as shown in the figure. If ratio of their radii is $1 : 2$ , and the ratio of the flux densities at the centre $O$ due to $A$ and $B$ is $1 : 3$ , then the value of $i_{1} / i_{2}$ will be
supporting img

1

\frac{1}{2}

2

\frac{1}{3}

3

\frac{1}{4}

4

\frac{1}{6}

Explanation:

$\frac{B_{1}}{B_{2}} = \frac{i_{1} r_{2}}{i_{2} r_{1}}$
$\frac{i_{1}}{i_{2}} = \frac{B_{1}}{B_{2}} \times \frac{r_{1}}{r_{2}} = \frac{1}{6}$

PHXII04:MOVING CHARGES AND MAGNETISM

362542 A circular coil connected to a battery of emf $E$ produces some magnetic field at its centre. The coil is unwound, stretched to double its length and rewound into a coil of $1 / 3 rd$ of the original radius and connected to a battery of emf $E$ ' such that magnetic field at its centre is same as earlier. Then $E^{'} =$

1

\frac{E}{2}

2

\frac{2 E}{9}

3

\frac{9 E}{4}

4

\frac{4 E}{9}

Explanation:

$B = n (\frac{μ_{0} i}{2 r}) \frac{μ_{0} n E}{2 r R}$
$(i = \frac{E}{R}, R i s t h e r e s i s t a n c e o f t h e w i r e)$
$E \propto \frac{r R}{n} H e r e r c h a n g e s t o \frac{r}{3}$
When the coil is streched its length increases but volume remains constant.
$l_{1} A_{1} = l_{2} A_{2} \Rightarrow l A = 2 l A_{2} (1)$
Where the new resistance is
$\frac{R_{1}}{R_{2}} = \frac{l_{1}}{A_{1}} \times \frac{A_{2}}{l_{2}}$
$= \frac{l_{1}}{A_{1}} \times (\frac{1}{2 l_{1}}) (\frac{l_{1} A_{1}}{l_{2}})$
$= \frac{l_{1}}{A_{1}} \times \frac{1}{2 l_{1}} \times \frac{l_{1} A_{1}}{2 l_{1}}$
$\frac{R_{1}}{R_{2}} = \frac{1}{4} \Rightarrow R_{2} = 4 R_{1}$
The ratio of number of turns is
$l_{2} = n_{2} 2 π (\frac{r}{3})$
$l_{1} = n_{1} 2 π r$
$\Rightarrow \frac{l_{2}}{l_{1}} = \frac{n_{2}}{3 n_{1}} = \frac{2 l_{1}}{l_{1}}$
$\Rightarrow n_{2} = 6 n_{1}$
$R$ changes to $4 R$ , $n$ chamges to $6 n$
$\frac{E}{E^{'}} = \frac{r R}{n} \times \frac{(6 n)}{\frac{r}{3} (4 R)} = \frac{9}{2}$
$\Rightarrow E^{'} = \frac{2 E}{9}$

PHXII04:MOVING CHARGES AND MAGNETISM

362543 In the given figure, magnetic field at centre is
supporting img

1

\frac{μ_{0} I}{8 R}

2 zero

3

\frac{μ_{0} I}{2 \sqrt{2} R}

4

\frac{μ_{0} I}{4 \sqrt{2} R}

Explanation:

$B_{n e t} = \sqrt{{(\frac{μ_{0} I}{4 R})}^{2} + {(\frac{μ_{0} I}{4 R})}^{2}}$
$= \frac{μ_{0} I}{4 R} \sqrt{2} = \frac{μ_{0} I}{2 \sqrt{2} R}$

PHXII04:MOVING CHARGES AND MAGNETISM

362540 A Helmholtz coil has a pair of loops, each with $N$ turns and radius $R$ . They are placed coaxially at distance $R$ and the same current $I$ flows through the loops in the same direction. The magnitude of magnetic field at $P$ , midway between the centres $A$ and $C$ , is given by [Refer to figure given below]:
supporting img

1

\frac{8 N μ_{0} I}{5^{1 / 2} R}

2

\frac{4 N μ_{0} I}{5^{3 / 2} R}

3

\frac{4 N μ_{0} I}{5^{1 / 2} R}

4

\frac{8 N μ_{0} I}{5^{3 / 2} R}

Explanation:

The resultant magnetic field at $P$ is
$B = 2 (\frac{μ_{0} N I R^{2}}{2 {(R^{2} + \frac{R^{2}}{4})}^{3 / 2}}) = \frac{8 μ_{0} N I}{5^{3 / 2} R}$

JEE - 2018

PHXII04:MOVING CHARGES AND MAGNETISM

362541 $A$ and $B$ are two concentric circular loop carrying current $i_{1}$ and $i_{2}$ , as shown in the figure. If ratio of their radii is $1 : 2$ , and the ratio of the flux densities at the centre $O$ due to $A$ and $B$ is $1 : 3$ , then the value of $i_{1} / i_{2}$ will be
supporting img

1

\frac{1}{2}

2

\frac{1}{3}

3

\frac{1}{4}

4

\frac{1}{6}

Explanation:

$\frac{B_{1}}{B_{2}} = \frac{i_{1} r_{2}}{i_{2} r_{1}}$
$\frac{i_{1}}{i_{2}} = \frac{B_{1}}{B_{2}} \times \frac{r_{1}}{r_{2}} = \frac{1}{6}$

PHXII04:MOVING CHARGES AND MAGNETISM

362542 A circular coil connected to a battery of emf $E$ produces some magnetic field at its centre. The coil is unwound, stretched to double its length and rewound into a coil of $1 / 3 rd$ of the original radius and connected to a battery of emf $E$ ' such that magnetic field at its centre is same as earlier. Then $E^{'} =$

1

\frac{E}{2}

2

\frac{2 E}{9}

3

\frac{9 E}{4}

4

\frac{4 E}{9}

Explanation:

$B = n (\frac{μ_{0} i}{2 r}) \frac{μ_{0} n E}{2 r R}$
$(i = \frac{E}{R}, R i s t h e r e s i s t a n c e o f t h e w i r e)$
$E \propto \frac{r R}{n} H e r e r c h a n g e s t o \frac{r}{3}$
When the coil is streched its length increases but volume remains constant.
$l_{1} A_{1} = l_{2} A_{2} \Rightarrow l A = 2 l A_{2} (1)$
Where the new resistance is
$\frac{R_{1}}{R_{2}} = \frac{l_{1}}{A_{1}} \times \frac{A_{2}}{l_{2}}$
$= \frac{l_{1}}{A_{1}} \times (\frac{1}{2 l_{1}}) (\frac{l_{1} A_{1}}{l_{2}})$
$= \frac{l_{1}}{A_{1}} \times \frac{1}{2 l_{1}} \times \frac{l_{1} A_{1}}{2 l_{1}}$
$\frac{R_{1}}{R_{2}} = \frac{1}{4} \Rightarrow R_{2} = 4 R_{1}$
The ratio of number of turns is
$l_{2} = n_{2} 2 π (\frac{r}{3})$
$l_{1} = n_{1} 2 π r$
$\Rightarrow \frac{l_{2}}{l_{1}} = \frac{n_{2}}{3 n_{1}} = \frac{2 l_{1}}{l_{1}}$
$\Rightarrow n_{2} = 6 n_{1}$
$R$ changes to $4 R$ , $n$ chamges to $6 n$
$\frac{E}{E^{'}} = \frac{r R}{n} \times \frac{(6 n)}{\frac{r}{3} (4 R)} = \frac{9}{2}$
$\Rightarrow E^{'} = \frac{2 E}{9}$

PHXII04:MOVING CHARGES AND MAGNETISM

362543 In the given figure, magnetic field at centre is
supporting img

1

\frac{μ_{0} I}{8 R}

2 zero

3

\frac{μ_{0} I}{2 \sqrt{2} R}

4

\frac{μ_{0} I}{4 \sqrt{2} R}

Explanation:

$B_{n e t} = \sqrt{{(\frac{μ_{0} I}{4 R})}^{2} + {(\frac{μ_{0} I}{4 R})}^{2}}$
$= \frac{μ_{0} I}{4 R} \sqrt{2} = \frac{μ_{0} I}{2 \sqrt{2} R}$

PHXII04:MOVING CHARGES AND MAGNETISM

362540 A Helmholtz coil has a pair of loops, each with $N$ turns and radius $R$ . They are placed coaxially at distance $R$ and the same current $I$ flows through the loops in the same direction. The magnitude of magnetic field at $P$ , midway between the centres $A$ and $C$ , is given by [Refer to figure given below]:
supporting img

1

\frac{8 N μ_{0} I}{5^{1 / 2} R}

2

\frac{4 N μ_{0} I}{5^{3 / 2} R}

3

\frac{4 N μ_{0} I}{5^{1 / 2} R}

4

\frac{8 N μ_{0} I}{5^{3 / 2} R}

Explanation:

The resultant magnetic field at $P$ is
$B = 2 (\frac{μ_{0} N I R^{2}}{2 {(R^{2} + \frac{R^{2}}{4})}^{3 / 2}}) = \frac{8 μ_{0} N I}{5^{3 / 2} R}$

JEE - 2018

PHXII04:MOVING CHARGES AND MAGNETISM

362541 $A$ and $B$ are two concentric circular loop carrying current $i_{1}$ and $i_{2}$ , as shown in the figure. If ratio of their radii is $1 : 2$ , and the ratio of the flux densities at the centre $O$ due to $A$ and $B$ is $1 : 3$ , then the value of $i_{1} / i_{2}$ will be
supporting img

1

\frac{1}{2}

2

\frac{1}{3}

3

\frac{1}{4}

4

\frac{1}{6}

Explanation:

$\frac{B_{1}}{B_{2}} = \frac{i_{1} r_{2}}{i_{2} r_{1}}$
$\frac{i_{1}}{i_{2}} = \frac{B_{1}}{B_{2}} \times \frac{r_{1}}{r_{2}} = \frac{1}{6}$

PHXII04:MOVING CHARGES AND MAGNETISM

362542 A circular coil connected to a battery of emf $E$ produces some magnetic field at its centre. The coil is unwound, stretched to double its length and rewound into a coil of $1 / 3 rd$ of the original radius and connected to a battery of emf $E$ ' such that magnetic field at its centre is same as earlier. Then $E^{'} =$

1

\frac{E}{2}

2

\frac{2 E}{9}

3

\frac{9 E}{4}

4

\frac{4 E}{9}

Explanation:

$B = n (\frac{μ_{0} i}{2 r}) \frac{μ_{0} n E}{2 r R}$
$(i = \frac{E}{R}, R i s t h e r e s i s t a n c e o f t h e w i r e)$
$E \propto \frac{r R}{n} H e r e r c h a n g e s t o \frac{r}{3}$
When the coil is streched its length increases but volume remains constant.
$l_{1} A_{1} = l_{2} A_{2} \Rightarrow l A = 2 l A_{2} (1)$
Where the new resistance is
$\frac{R_{1}}{R_{2}} = \frac{l_{1}}{A_{1}} \times \frac{A_{2}}{l_{2}}$
$= \frac{l_{1}}{A_{1}} \times (\frac{1}{2 l_{1}}) (\frac{l_{1} A_{1}}{l_{2}})$
$= \frac{l_{1}}{A_{1}} \times \frac{1}{2 l_{1}} \times \frac{l_{1} A_{1}}{2 l_{1}}$
$\frac{R_{1}}{R_{2}} = \frac{1}{4} \Rightarrow R_{2} = 4 R_{1}$
The ratio of number of turns is
$l_{2} = n_{2} 2 π (\frac{r}{3})$
$l_{1} = n_{1} 2 π r$
$\Rightarrow \frac{l_{2}}{l_{1}} = \frac{n_{2}}{3 n_{1}} = \frac{2 l_{1}}{l_{1}}$
$\Rightarrow n_{2} = 6 n_{1}$
$R$ changes to $4 R$ , $n$ chamges to $6 n$
$\frac{E}{E^{'}} = \frac{r R}{n} \times \frac{(6 n)}{\frac{r}{3} (4 R)} = \frac{9}{2}$
$\Rightarrow E^{'} = \frac{2 E}{9}$

PHXII04:MOVING CHARGES AND MAGNETISM

362543 In the given figure, magnetic field at centre is
supporting img

1

\frac{μ_{0} I}{8 R}

2 zero

3

\frac{μ_{0} I}{2 \sqrt{2} R}

4

\frac{μ_{0} I}{4 \sqrt{2} R}

Explanation:

$B_{n e t} = \sqrt{{(\frac{μ_{0} I}{4 R})}^{2} + {(\frac{μ_{0} I}{4 R})}^{2}}$
$= \frac{μ_{0} I}{4 R} \sqrt{2} = \frac{μ_{0} I}{2 \sqrt{2} R}$

PHXII04:MOVING CHARGES AND MAGNETISM

362540 A Helmholtz coil has a pair of loops, each with $N$ turns and radius $R$ . They are placed coaxially at distance $R$ and the same current $I$ flows through the loops in the same direction. The magnitude of magnetic field at $P$ , midway between the centres $A$ and $C$ , is given by [Refer to figure given below]:
supporting img

1

\frac{8 N μ_{0} I}{5^{1 / 2} R}

2

\frac{4 N μ_{0} I}{5^{3 / 2} R}

3

\frac{4 N μ_{0} I}{5^{1 / 2} R}

4

\frac{8 N μ_{0} I}{5^{3 / 2} R}

Explanation:

The resultant magnetic field at $P$ is
$B = 2 (\frac{μ_{0} N I R^{2}}{2 {(R^{2} + \frac{R^{2}}{4})}^{3 / 2}}) = \frac{8 μ_{0} N I}{5^{3 / 2} R}$

JEE - 2018

PHXII04:MOVING CHARGES AND MAGNETISM

362541 $A$ and $B$ are two concentric circular loop carrying current $i_{1}$ and $i_{2}$ , as shown in the figure. If ratio of their radii is $1 : 2$ , and the ratio of the flux densities at the centre $O$ due to $A$ and $B$ is $1 : 3$ , then the value of $i_{1} / i_{2}$ will be
supporting img

1

\frac{1}{2}

2

\frac{1}{3}

3

\frac{1}{4}

4

\frac{1}{6}

Explanation:

$\frac{B_{1}}{B_{2}} = \frac{i_{1} r_{2}}{i_{2} r_{1}}$
$\frac{i_{1}}{i_{2}} = \frac{B_{1}}{B_{2}} \times \frac{r_{1}}{r_{2}} = \frac{1}{6}$

PHXII04:MOVING CHARGES AND MAGNETISM

362542 A circular coil connected to a battery of emf $E$ produces some magnetic field at its centre. The coil is unwound, stretched to double its length and rewound into a coil of $1 / 3 rd$ of the original radius and connected to a battery of emf $E$ ' such that magnetic field at its centre is same as earlier. Then $E^{'} =$

1

\frac{E}{2}

2

\frac{2 E}{9}

3

\frac{9 E}{4}

4

\frac{4 E}{9}

Explanation:

$B = n (\frac{μ_{0} i}{2 r}) \frac{μ_{0} n E}{2 r R}$
$(i = \frac{E}{R}, R i s t h e r e s i s t a n c e o f t h e w i r e)$
$E \propto \frac{r R}{n} H e r e r c h a n g e s t o \frac{r}{3}$
When the coil is streched its length increases but volume remains constant.
$l_{1} A_{1} = l_{2} A_{2} \Rightarrow l A = 2 l A_{2} (1)$
Where the new resistance is
$\frac{R_{1}}{R_{2}} = \frac{l_{1}}{A_{1}} \times \frac{A_{2}}{l_{2}}$
$= \frac{l_{1}}{A_{1}} \times (\frac{1}{2 l_{1}}) (\frac{l_{1} A_{1}}{l_{2}})$
$= \frac{l_{1}}{A_{1}} \times \frac{1}{2 l_{1}} \times \frac{l_{1} A_{1}}{2 l_{1}}$
$\frac{R_{1}}{R_{2}} = \frac{1}{4} \Rightarrow R_{2} = 4 R_{1}$
The ratio of number of turns is
$l_{2} = n_{2} 2 π (\frac{r}{3})$
$l_{1} = n_{1} 2 π r$
$\Rightarrow \frac{l_{2}}{l_{1}} = \frac{n_{2}}{3 n_{1}} = \frac{2 l_{1}}{l_{1}}$
$\Rightarrow n_{2} = 6 n_{1}$
$R$ changes to $4 R$ , $n$ chamges to $6 n$
$\frac{E}{E^{'}} = \frac{r R}{n} \times \frac{(6 n)}{\frac{r}{3} (4 R)} = \frac{9}{2}$
$\Rightarrow E^{'} = \frac{2 E}{9}$

PHXII04:MOVING CHARGES AND MAGNETISM

362543 In the given figure, magnetic field at centre is
supporting img

1

\frac{μ_{0} I}{8 R}

2 zero

3

\frac{μ_{0} I}{2 \sqrt{2} R}

4

\frac{μ_{0} I}{4 \sqrt{2} R}

Explanation:

$B_{n e t} = \sqrt{{(\frac{μ_{0} I}{4 R})}^{2} + {(\frac{μ_{0} I}{4 R})}^{2}}$
$= \frac{μ_{0} I}{4 R} \sqrt{2} = \frac{μ_{0} I}{2 \sqrt{2} R}$

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